I considered deleting the post, but this seems more cowardly than just admitting I was wrong. But TIL something!
This kind of thread is why I duck out of casual maths discussions as a maths PhD.
The two sets have the same value, that is the value of both sets is unbounded. The set of 100s approaches that value 100 times quicker than the set of singles. Completely intuitive to someone who’s taken first year undergraduate logic and calculus courses. Completely unintuitive to the lay person, and 100 lay people can come up with 100 different wrong conclusions based on incorrect rationalisations of the statement.
I’ve made an effort to just not participate since back when people were arguing Rick and Morty infinite universe bollocks. “Infinite universes means there are an infinite number of identical universes” really boils my blood.
It’s why I just say “algebra” when asked what I do. Even explaining my research (representation theory) to a tangentially related person, like a mathematical physicist, just ends in tedious non-discussion based on an incorrect framing of my work through their lens of understanding.
For what it’s worth, people actually taking the time to explain helped me see the error in my reasoning.
There’s no problem at all with not understanding something, and I’d go so far as to say it’s virtuous to seek understanding. I’m talking about a certain phenomenon that is overrepresented in STEM discussions, of untrained people (who’ve probably took some internet IQ test) thinking they can hash out the subject as a function of raw brainpower. The ceiling for “natural talent” alone is actually incredibly low in any technical subject.
There’s nothing wrong with meming on a subject you’re not familiar with, in fact it’s often really funny. It’s the armchair experts in the thread trying to “umm actually…” the memer when their “experience” is a YouTube video at best.
I don’t know why you see it as an error. It’s the format of the meme. The guy in the middle is right, the guy on the left is wrong. That’s just how this meme works. But the punchline in this meme format is the the guy on the right agrees with the wrong guy in an unexpected way. I’m with the guy on the right and no appeals to Schröder–Bernstein theorem is going to change my mind.
Yeah I sell cabinets and sometimes people are like “How much would a 24 inch cabinet cost?”
It could cost anything!
Then there are customers like “It’s the same if I just order them online right?” and I say “I wouldn’t recommend it. There’s a lot of little details to figure out and our systems can be error probe anyway…” then a month later I’m dealing with an angry customer who ordered their stuff online and is now mad at me for stuff going wrong.
The two sets have the same value, that is the value of both sets is unbounded. The set of 100s approaches that value 100 times quicker than the set of singles.
Hey. Sorry, I’m not at all a mathematician, so this is fascinating to me. Doesn’t this mean that, once the two sets have reached their value, the set of 100 dolar bills will weigh 100 times less (since both bills weigh the same, and there are 100 times fewer of one set than the other)?
If so, how does it reconcile with the fact that there should be the same number bills in the sets, therefore the same weight?
I like this comment. It reads like a mathematician making a fun troll based on comparing rates of convergence (well, divergence considering the sets are unbounded). If you’re not a mathematician, it’s actually a really insightful comment.
So the value of the two sets isn’t some inherent characteristic of the two sets. It is a function which we apply to the sets. Both sets are a collection of bills. To the set of singles we assign one value function: “let the value of this set be $1 times the number of bills in this set.” To the set of hundreds we assign a second value function: “let the value of this set be $100 times the number of bills in this set.”
Now, if we compare the value restricted to two finite subsets (set within a set) of the same size, the subset of hundreds is valued at 100 times the subset of singles.
Comparing the infinite set of bills with the infinite set of 100s, there is no such difference in values. Since the two sets have unbounded size (i.e. if we pick any number N no matter how large, the size of these sets is larger) then naturally, any positive value function applied to these sets yields an unbounded number, no mater how large the value function is on the hundreds “I decide by fiat that a hundred dollar bill is worth $1million” and how small the value function is on the singles “I decide by fiat that a single is worth one millionth of a cent.”
In overly simplified (and only slightly wrong) terms, it’s because the sizes of the sets are so incalculably large compared to any positive value function, that these numbers just get absorbed by the larger number without perceivably changing anything.
The weight question is actually really good. You’ve essentially stumbled upon a comparison tool which is comparing the rates of convergence. As I said previously, comparing the value of two finite subsets of bills of the same size, we see that the value of the subset of hundreds is 100 times that of the subset of singles. This is a repeatable phenomenon no matter what size of finite set we choose. By making a long list of set sizes and values “one single is worth $1, 2 singles are worth $2,…” we can define a series which we can actually use for comparison reasons. Note that the next term in the series of hundreds always increases at a rate of 100 times that of the series of singles. Using analysis techniques, we conclude that the set of hundreds is approaching its (unbounded) limit at 100 times the rate of the singles.
The reason we cannot make such comparisons for the unbounded sets is that they’re unbounded. What is the weight of an unbounded number of hundreds? What is the weight of an unbounded number of collections of 100x singles?
Is it possible for infinite numbers to be larger than others? Or are all infinite numbers equal?
once the two sets have reached their value
will weigh 100 times less
there should be the same number bills in the sets
The short answer is that none of these statements apply the way you think to infinite sets.
So to paraphrase, the raging person in the middle is right? I’ll take your answer no questions asked.
If we only consider the monetary value, both “briefs” have the same value. Otherwise if we incorporate utility theory with a concave bounded utility curve over the monetary value and factor in other terms such as ease of payments, or weight (of the drawn money) then the “worth” of the 100 dollar bills brief could be greater for some people. For me, the 1 dollar bills brief has more value since I’m considering a potential tax evasion prosecution. It would be very suspicious if I go around paying everything with 100 dollar bills, whereas there’s a limit on my daily spending with the other brief (how many dollars I can count out of the brief and then handle to the other person).
I admit the only time I’ve encountered the word utility as an algebraist is when I had to TA Linear Optimisation & Game Theory; it was in the sections of notes for the M level course that wasn’t examinable for the Bachelors students so I didn’t bother reading it. My knowledge caps out at equilibria of mixed strategies. It’s interesting to see that there’s some rigorous way of codifying user preference. I’ll have to read about it at some point.
Correct me if I’m wrong, but isn’t it that a simple statement(this is more worth than the other) can’t be done, since it isn’t stated how big the infinities are(as example if the 1$ infinity is 100 times bigger they are worth the same).
Sorry if you’ve seen this already, as your comment has just come through. The two sets are the same size, this is clear. This is because they’re both countably infinite. There isn’t such a thing as different sizes of countably infinite sets. Logic that works for finite sets (“For any finite a and b, there are twice as many integers between a and b as there are even integers between a and b, thus the set of integers is twice the set of even integers”) simply does not work for infinite sets (“The set of all integers has the same size as the set of all even integers”).
So no, it isn’t due to lack of knowledge, as we know logically that the two sets have the exact same size.
